Sec. 11.5 Vector methods in geometry 197
From this
s
1 −t
=
1 −v
1 −w
,
t
1 −r
=
1 −w
1 −u
,
r
1 −s
=
1 −u
1 −v
,
and so by multiplication
s
1 −t
t
1 −r
r
1 −s
= 1.
Thus we obtain our conclusion. This is known asC eva’s th eo rem.
In fact we also have that
u
s
=−
1 −u
1 −v
,
v
t
=−
1 −v
1 −w
,
w
r
=−
1 −w
1 −u
,
which givesuvw=−rst.Thisis
Z 0 Z 4
Z 0 Z 1
Z 0 Z 5
Z 0 Z 2
Z 0 Z 6
Z 0 Z 3
=−
Z 2 Z 4
Z 2 Z 3
Z 3 Z 5
Z 3 Z 1
Z 1 Z 6
Z 1 Z 2
.
CONVERSEofC eva’s th eo rem.Conversely, for non-collinear points Z 1 ,Z 2
and Z 3 ,letZ 4 ∈Z 2 Z 3 ,Z 5 ∈Z 3 Z 1 and Z 6 ∈Z 1 Z 2. If (11.5.1) holds and Z 2 Z 5 and Z 3 Z 6
meet at a point Z 0 ,thenZ 1 Z 4 also passes through Z 0.
To start our proof we note that we have
Z 5 =( 1 −v)Z 0 +vZ 2 =( 1 −s)Z 3 +sZ 1 ,
Z 6 =( 1 −w)Z 0 +wZ 3 =( 1 −t)Z 1 +tZ 2.
Hence
Z 0 =
s
1 −v
Z 1 −
v
1 −v
Z 2 +
1 −s
1 −v
Z 3 ,Z 0 =
1 −t
1 −w
Z 1 +
t
1 −w
Z 2 −
w
1 −w
Z 3.
It follows that
s
1 −v
=
1 −t
1 −w
,−
v
1 −v
=
t
1 −w
,
1 −s
1 −v
=−
w
1 −w
,
from which
s
1 −t
=
1 −v
1 −w
,( 1 −s)t=vw.
On eliminatingsbetween these, we obtain( 1 −v)t^2 +(v−w)t−vw( 1 −w)=0. We
then obtain two pairs of solutions,t=w,s= 1 −v,and
t=−v
1 −w
1 −v
,s=
1 −vw
1 −w
.
The first pair of solutions leads tov=w=0andsoZ 5 =Z 6 =Z 0 =Z 1 ,whichwe
regard as a degenerate case.
WithZ 4 =( 1 −r)Z 2 +rZ 3 , we are given that
1 −r
r
=
st
( 1 −s)( 1 −t)